Problem 42 - Algebra for College Students

Question

Solve each equation. Don't forget to check each of your potential solutions. \(\sqrt{-x}-6=x\)

Step-by-Step Solution

Answer

The solution is \(x = -4\).

Step 1: Move X to One Side

First, we want to isolate the square root on one side of the equation. We have \(\sqrt{-x} - 6 = x\). Let's move x to one side of the equation by adding x to both sides. This will result in \(\sqrt{-x} = x + 6\).

Step 2: Square Both Sides

To eliminate the square root, we square both sides of the equation. After squaring, we get \(-x = (x + 6)^2\).

Step 3: Expand and Simplify

Next, we expand \((x + 6)^2\), which results in \(x^2 + 12x + 36\). So the equation now becomes \(-x = x^2 + 12x + 36\).

Step 4: Rearrange the Equation

To make the equation solvable, add x to both sides to get the equation in the form of a standard quadratic equation: \(0 = x^2 + 13x + 36\).

Step 5: Solve the Quadratic Equation

To solve \(x^2 + 13x + 36 = 0\), we can factor the quadratic equation. It factors into \((x+4)(x+9) = 0\).

Step 6: Find Potential Solutions

Set each factor equal to zero: \(x + 4 = 0\) or \(x + 9 = 0\). This gives us \(x = -4\) or \(x = -9\) as potential solutions.

Step 7: Check Each Solution

Substitute \(x = -4\) back into the original equation: \(\sqrt{-(-4)} - 6 = -4\). This simplifies to \(\sqrt{4} - 6 = -4\), which is \(2 - 6 = -4\), a true statement. Now check \(x = -9\): \(\sqrt{-(-9)} - 6 = -9\) simplifies to \(\sqrt{9} - 6 = -9\), which is \(3 - 6 = -3\), a false statement. Thus, \(x = -4\) is the only valid solution.

Key Concepts

Solving Equations

When we solve equations, our goal is to find the value of the variable that makes the equation true. In the exercise given, the equation is \(\sqrt{-x} - 6 = x\). To solve it, we start by isolating the square root term on one side of the equation. This involves a series of steps that simplify the expression gradually. For this problem, we can achieve that by adding \(x\) to both sides, resulting in \(\sqrt{-x} = x + 6\). Next, to eliminate the square root symbol, we square both sides of the equation.

• This squaring process is essential when dealing with square roots, as it helps us to convert the equation into a quadratic form, which is typically more straightforward to solve.
• After squaring, the equation evolves into \(-x = (x + 6)^2\).

Recognizing this transformation is a critical skill in solving more complex algebraic equations.

Checking Solutions

Once you have potential solutions to an equation, you must check each one to confirm it's valid. For our exercise, the potential solutions after factoring are \(x = -4\) and \(x = -9\). It is crucial to substitute these values back into the original equation to ensure they satisfy the original conditions.

• Check \(x = -4\): Substituting \(-4\) back in gives us \(\sqrt{4} - 6 = -4\), which simplifies to \(2 - 6 = -4\). This is true, so \(x = -4\) is a valid solution.
• Check \(x = -9\): Substitute \(-9\) back into the original equation \(\sqrt{9} - 6 = -9\), which simplifies to \(3 - 6 = -3\). This isn't true, so \(x = -9\) is not valid.

This checking process helps us avoid false solutions that sometimes appear when squaring both sides of an equation.

Factoring Quadratics

Factoring quadratics is a method used to solve second-degree polynomial equations like \(x^2 + 13x + 36 = 0\). Factoring involves finding two binomials that multiply to give the original quadratic equation. Here, the quadratic factors into \((x+4)(x+9) = 0\).

• To factor, look for two numbers that multiply to the constant term (36 in this case) and add to the linear coefficient (13 here).
• Both factors of \(36\) that add up to \(13\) are \(4\) and \(9\).

Setting each factor equal to zero gives you the potential solutions: \(x = -4\) and \(x = -9\). This technique is a foundational skill in algebra, making it easier to handle a variety of equations.

Square Roots

Understanding square roots is crucial when solving equations involving them, such as \(\sqrt{-x} = x + 6\). Square roots "undo" squaring, and knowing how to manipulate them allows us to simplify complex expressions.

• A square root, denoted by \(\sqrt{}\), finds a number which, when multiplied by itself, gives the original number under the root.
• To isolate a square root term, as we did by moving terms around in the equation, helps set up further steps for solving.

Squaring both sides was necessary here to remove the root, allowing us to solve for \(x\) with fewer complexities. Recognizing when and how to deal with square roots is key for tackling higher-level math problems efficiently.